HDU 4507 吉哥系列故事 恨7不成妻 （数位dp）

题意:

中文题 题意很明显啦

分析:

题意要求平方和,但是做过类似线段树那种题目的人都知道不能直接维护的
可以通过cnt,sum,来算出sq
状态很明显嘛,dp[i][mod][sum]:=i位,数mod 7,数位和mod 7
cnt[i]=∑9d=0{cnt[d]}
sum[i]=∑9d=0{cnt[d]∗d∗10i−1+sum[d]}
sq[i]=∑9d=0{cnt[d]∗(d∗10i−1+xi−11)2}
       =∑9d=0{cnt[d]∗{(d∗10i−1)2+2∗(d∗10i−1)∗xi−11+xi−112}}
       =∑9d=0{cnt[d]∗(d∗10i−1)2+2∗(d∗10i−1)∗sum[d]+sq[d]}

代码:

////  Created by TaoSama on 2015-10-22//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;struct Node {    LL cnt, sum, sq;    Node(int x = -1) {cnt = sum = sq = x;}};Node dp[20][7][7];LL p[20], digit[20];Node dfs(int i, int mod, int sum, int e) {    Node &f = dp[i][mod][sum], suf, ret(0);    if(!i) {        ret.cnt = mod && sum;        return ret;    }    if(!e && ~f.cnt) return f;    int to = e ? digit[i] : 9;    for(int d = 0; d <= to; ++d) {        if(d == 7) continue;        suf = dfs(i - 1, (mod * 10 + d) % 7, (sum + d) % 7, e && d == to);        LL ten = p[i - 1];        ret.cnt = (ret.cnt + suf.cnt) % MOD;        LL b = d * ten % MOD, bsq = b * b % MOD;        ret.sum = (ret.sum + b * suf.cnt % MOD + suf.sum) % MOD;        ret.sq = (ret.sq + suf.cnt * bsq % MOD + 2 * b * suf.sum % MOD + suf.sq) % MOD;    }    return e ? ret : f = ret;}LL calc(LL x) {    int cnt = 0;    for(; x; x /= 10) digit[++cnt] = x % 10;    return dfs(cnt, 0, 0, 1).sq;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    p[0] = 1;    for(int i = 1; i < 20; ++i) p[i] = p[i - 1] * 10 % MOD;    while(t--) {        LL l, r; scanf("%I64d%I64d", &l, &r);        printf("%I64d\n", (calc(r) - calc(l - 1) + MOD) % MOD);    }    return 0;}